python3.x 基础三：set集合

集合，set()，记住：

1个特点：去重，把列表变成集合，达到自动去重操作，无序

5个关系：测试两个列表的交差并子反向差集

方法：

•  |  add(...) 常用，已存在元素去重不生效
•  |      Add an element to a set.
•  |     This has no effect if the element is already present.

  >>> list1=[3,2,1,1,2,3,4,5]
  >>> set(list1)
  {1, 2, 3, 4, 5}
  >>> list2=[3,4,5,6,7,8]
  >>> set(list1).add(2)
  >>> set(list1).add(6)
  >>> print(set(list1).add(2))
  None
  >>> print(set(list1).add(6))
  None
  >>> set1=set(list1)
  >>> set2=set(list2)
  >>> set1,set2
  ({1, 2, 3, 4, 5}, {3, 4, 5, 6, 7, 8})
  >>> set1.add(3)
  >>> print(set1.add(3)
  ... )
  None
  >>> print(set1.add(7))
  None
  >>> set1.add('aaa')
  >>> set1
  {1, 2, 3, 4, 5, 7, 'aaa'}
  >>> id(set1)
  140138768484616
  >>> set1.add('aaaa')
  >>> id(set1)
  140138768484616
  >>> set1
  {1, 2, 3, 4, 5, 7, 'aaa', 'aaaa'}
  >>> set1.add('7')
  >>> set1
  {1, 2, 3, 4, 5, 7, 'aaa', 'aaaa', '7'}
  >>> set1.add(7)
  >>> set1
  {1, 2, 3, 4, 5, 7, 'aaa', 'aaaa', '7'}

•  如果是字符串，则拆分成单个字符集合

• >>> set('abc')
  {'a', 'c', 'b'}

   

 

•  |  clear(...) 清空一个集合
•  |      Remove all elements from this set.

  >>> set1.clear()
  >>> set1
  set()

   

•  |  copy(...) 影子复制，指向同一个内存地址
•  |      Return a shallow copy of a set. |  

  >>> list1
  [3, 2, 1, 1, 2, 3, 4, 5]
  >>> list2
  [3, 4, 5, 6, 7, 8]
  >>> set1=set(list1)
  >>> id(set1)
  140138768485512>>> set3=set1.copy()
  >>> id(set3)
  140138695576712

   

•  |  difference(...) 差集，格式set1.difference(set2)，求in list1 not in list2的集合
•  |      Return the difference of two or more sets as a new set. |   
•  |      (i.e. all elements that are in this set but not the others.)   

  >>> set1
  {1, 2, 3, 4, 5}
  >>> set2
  {3, 4, 5, 6, 7, 8}
  >>> set1.difference(set2)
  {1, 2}

   

•  |  difference_update(...) 删除在本集合同时也在其他集合的元素，差集
•  |      Remove all elements of another set from this set. |

  >>> set1=set(list1)
  >>> set2=set(list2)
  >>> set1,set2
  ({1, 2, 3, 4, 5}, {3, 4, 5, 6, 7, 8})
  >>> set1=set(list1)
  >>> set2=set(list2)
  >>> set2.difference_update(set1)
  >>> set2
  {6, 7, 8}

   

   
•  |  discard(...) 删除一个在本集合中的元素
•  |      Remove an element from a set if it is a member.
•  |      
•  |      If the element is not a member, do nothing. |  

  >>> set3=set([1,2,3,'a','b','c'])
  >>> set3.discard(1)
  >>> set3
  {2, 3, 'c', 'b', 'a'}
  >>> set3.discard('a')
  >>> set3
  {2, 3, 'c', 'b'}
  >>> set3.discard('dd')
  >>> set3
  {2, 3, 'c', 'b'}

   

•  |  intersection(...)并集，同时在两个集合中的元素
•  |      Return the intersection of two sets as a new set.
•  |      
•  |      (i.e. all elements that are in both sets.) |  

  >>> set1
  {1, 2, 3, 4, 5}
  >>> set2
  {3, 4, 5, 6, 7, 8}
  >>> set1.intersection(set2)
  {3, 4, 5}
  >>> set2.intersection(set1)
  {3, 4, 5}

•  |  intersection_update(...) 交集
•  |      Update a set with the intersection of itself and another. |  

  >>> set1,set2
  ({1, 2, 3, 4, 5}, {3, 4, 5, 6, 7, 8})
  >>> set1.intersection_update(set2)
  >>> set1
  {3, 4, 5}

•  |  isdisjoint(...) 返回布尔值，判断两个集合是否没有并集
•  |      Return True if two sets have a null intersection.
•  |  

  >>> set1,set2,set3,set4
  ({3, 4, 5}, {3, 4, 5, 6, 7, 8}, {2, 3, 'c', 'b'}, {'y', 'x', 'z'})
  >>> set1.isdisjoint(set2)
  False
  >>> set1.isdisjoint(set4)
  True

•  |  issubset(...) 返回布尔值，判断前一个集合是否是后一个集合的子集
•  |      Report whether another set contains this set. |  

  >>> set1
  {3, 4, 5}
  >>> set5
  {3, 4}
  >>> set5.issubset(set1)
  True

•  |  issuperset(...) 返回布尔值，判断前一个集合是否是后一个集合的父集
•  |      Report whether this set contains another set. |  

  >>> set1
  {3, 4, 5}
  >>> set5
  {3, 4}
  >>> set1.issuperset(set5)
  True

•  |  pop(...) 随机删除一个集合元素，返回被删除元素，空集合删除则报错
•  |      Remove and return an arbitrary set element.
•  |      Raises KeyError if the set is empty. |  

  >>> set1
  {3, 4, 5}
  >>> set1.pop()
  3
  >>> set1
  {4, 5}
  >>> set1.pop()
  4
  >>> set1.pop()
  5
  >>> set1.pop()
  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
  KeyError: 'pop from an empty set'

•  |  remove(...) 删除指定在集合中的元素
•  |      Remove an element from a set; it must be a member. |      

  >>> set1=set(list1)
  >>> set1
  {1, 2, 3, 4, 5}
  >>> set1.remove(1)
  >>> set1
  {2, 3, 4, 5}
  >>> set1.remove('a')
  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
  KeyError: 'a'

•  |  symmetric_difference(...) 对称差集，集合A与集合B不相交的部分，交集的反集
•  |     Return the symmetric difference of two sets as a new set.
•  |     (i.e. all elements that are in exactly one of the sets.)     
  

  >>> set1
  {1, 2, 3, 4, 5}
  >>> set6={4,5,6,7,8}
  >>> set1.symmetric_difference(set6)
  {1, 2, 3, 6, 7, 8}
  >>> set6.symmetric_difference(set1)
  {1, 2, 3, 6, 7, 8}
  >>> set1,set7
  ({1, 2, 3, 4, 5}, {'a', 'c', 'b'})
  >>> set1.symmetric_difference(set7)
  {'c', 2, 3, 1, 4, 5, 'b', 'a'}

•  |  symmetric_difference_update(...)
•  |      Update a set with the symmetric difference of itself and another.
•  |  

•  |  union(...) 并集
•  |      Return the union of sets as a new set.
•  |      (i.e. all elements that are in either set.)
  

  >>> set1
  {1, 2, 3, 4, 5}
  >>> set2
  {3, 4, 5, 6, 7, 8}
  >>> set1.union(set2)
  {1, 2, 3, 4, 5, 6, 7, 8}

•  |  update(...) 用交集更新到set1的集合
•  |      Update a set with the union of itself and others. |  

  >>> set1
  {1, 2, 3, 4, 5}
  >>> set2
  {3, 4, 5, 6, 7, 8}
  >>> set1.update(set2)
  >>> set1
  {1, 2, 3, 4, 5, 6, 7, 8}
  >>>